30
31
2-misol.
Funksiyaning hosilasini toping
(
k, b –
o‘zgar mas sonlar):
1)
f
(
x
) = (
kx
+
b
)
n
;
2)
f
(
x
)
=
sin(
kx + b
);
3)
f
(
x
)
=
cos(
kx + b
);
4)
f
(
x
)
=
tg(
kx + b
)
.
1)
f
(
t
)
=
t
n
va
t
(
x
)
= kx + b
funksiyalarga (1) formulani qo‘llaymiz:
((
kx
+
b
)
n
)′ = (
t
n
)′
∙
(
kx+b
)′
= n
t
n–
1
∙k = n∙k∙
(
kx + b
)
n
–1
.
2)
f
(
t
)
=
sin
t
va
t
(
x
)
=kx+b
funksiyalarga (1) formulani qo‘llaymiz:
(sin(
kx+b
))′ = (sin
t
)′
∙
(
kx+b
)′
= k ∙
cos
t = k ∙
cos(
kx + b
)
.
3)
f
(
t
)
=
cos
t
va
t
(
x
)
= kx + b
funksiyalarga (1) formulani qo‘llaymiz:
(cos(
kx + b
))′
=
(cos
t
)′
∙
(
kx+b
)′
= –k ∙
sin
t = – k ∙
sin(
kx + b
)
.
4)
f
(
t
)
=
tg
t
va
t
(
x
)
= kx+b
funksiyalarga (1) formulani qo‘llaymiz:
(tg(
kx + b
))
′
=
(tg
t
)
′
∙
(
kx + b
)
′
(
)
2
2
1
cos
cos
k
k
t
kx b
=
⋅ =
+
.
Javob:
1) ((
kx
+
b
)
n
)′
=
n∙k∙
(
kx + b
)
n
–1
;
2) (sin(
kx + b
))
′
= k∙
cos(
kx+b
);
3)
(cos(
kx+b
)
′
=–k∙
sin(
kx+b
); 4)
(tg(
kx+b
)
′
=
(
)
(
)
'
2
(
cos
k
tg kx b
kx b
+
=
+
.
▲
3-misol.
f
(
x
)
=
sin8
x∙e
(3
x
+2)
funksiya hosilasini toping.
Hosilani topishning 4-qoidasi hamda (1) formulani qo‘llab
hosilani
topamiz:
f
′
(
x
) = (sin8
xe
(3
x
+2)
)
′
= (sin8
x
)
′
e
3
x
+2
+ sin8
x
∙ (
e
3
x
+2
)
′
= cos8
xe
3
x
+2
∙
(8
x
)
′
+
+ sin8
xe
3
x
+2
∙
(3
x+
2)
′
=
e
3
x
+2
∙
(8cos8
x
+ 3sin8
x
).
Javob:
e
3
x
+2
∙
(8cos8
x
+ 3sin8
x
).
▲
4-misol.
h
(
x
) = (
x
3
+ 1)
5
funksiyaning
x
0
= 1 nuqtadagi hosilasini toping.
(1) formuladan foydalanib hosilani hisoblaymiz:
h
′
(
x
) = 5(
x
3
+1)
4
(
x
3
+1)
′
= 5(
x
3
+1)
4
3
x
2
= 15
x
2
(
x
3
+1)
4
.
Demak,
h
′
(1) = 15(1
3
+1)
4
·1
2
= 15·16 = 240.
Javob
: 240.
▲
5-misol.
f
(
x
) = 2
cos
x
funksiyaning hosilasini toping.
(1) formuladan foydalanib hosilani hisoblaymiz:
f
′
(
x
) =2
cos
x
ln2 (cos
x
)
′
= –sin
x
2
cos
x
ln2.
Javob
: – sin
x
2
cos
x
ln2.
▲
32
33
6-misol.
f
(
x
) = tg
5
x
funksiyaning hosilasini toping.
(1) formuladan foydalanib hosilani hisoblaymiz:
f
′
(
x
) =5tg
4
x
(tg
x
)
′ = 5tg
4
x
1
2
cos
x
.
Javob
:
4
2
5tg .
cos
x
x
▲
7-misol.
h
(
x
)=3
cos
x
·log
7
(
x
3
+2
x
) funksiyaning hosilasini toping.
f
(
x
)=3
cos
x
va
g
(
x
)=log
7
(
x
3
+2
x
) belgilashlarni kiritib, (1) formulani –
murakkab funksiya hosilasini topish formulasini qo‘llaymiz:
f
ʹ(
x
)=(3
cos
x
)ʹ=3
cos
x
ln3·(cos
x
)ʹ
=–3
cos
x
ln3·sin
x
,
g
ʹ(
x
)=(log
7
(
x
3
+2
x
))ʹ=
3
1
(
2 )ln 7
x
x
+
·(
x
3
+2
x
)ʹ=
2
3
3
2
(
2 )ln 7
x
x
x
+
+
hamda
h
(
x
) funksiyani 2 ta funksiyaning ko‘paytmasi deb qaraymiz:
h
ʹ(
x
) = (3
cos
x
∙
log
7
(
x
3
+2
x
))ʹ = (3
cos
x
)ʹ·log
7
(
x
3
+2
x
) +
+ 3
cos
x
⋅
(log
7
(
x
3
+2
x
))ʹ
= –3
cos
x
·ln3·sin
x
·log
7
(
x
3
+2
x
)
+
cos
2
3
3 (3
2)
(
2 )ln 7
x
x
x
x
+
+
.
Javob
: –3
cos
x
·ln3·sin
x
·log
7
(
x
3
+2
x
)
+
cos
2
3
3 (3
2)
(
2 )ln 7
x
x
x
x
+
+
.
▲
?
Savol va topshiriqlar
1. Murakkab funksiya deb nimaga aytiladi? Misol keltiring.
2. Murakkab funksiyaning aniqlanish sohasi qanday topiladi?
3. Murakkab funksiya hosilasini topish formulasini yoza olasizmi?
4. Murakkab funksiya hosilasini topishni 1–2 ta misolda ko‘rsating.
32
33
Mashqlar
35.
Agar
1
)
(
2
−
=
x
x
f
bo‘lsa, ko‘rsatilgan funksiyalarni toping:
1)
)
1
(
x
f
; 2)
);
2
(
x
f
)
; 3)
);
1
(
2
−
x
f
)
; 4)
).
1
(
)
1
(
−
−
+
x
f
x
f
)
.
36.
Agar
1
1
)
(
−
+
=
x
x
x
f
bo‘lsa, ko‘rsatilgan funksiyalarni toping:
1)
)
1
(
x
f
; 2)
)
1
(
2
x
f
; 3)
);
1
(
−
x
f
)
; 4)
)
1
(
+
x
f
.
37.
Agar
1
)
(
,
)
(
2
−
=
=
x
x
g
x
x
f
bo‘lsa, quyidagilarni toping:
1)
));
(
(
x
g
f
2)
));
(
(
x
f
f
3)
));
(
(
x
g
g
4)
)).
(
(
x
f
g
38
. Agar
1
)
(
,
)
(
2
3
+
=
=
x
x
g
x
x
f
bo‘lsa, quyidagilarni toping:
1)
;
1
)
(
)
(
2
−
x
g
x
f
2)
2
3
)
(
3
)
(
−
+
+
x
x
g
x
f
;
3)
));
(
(
x
g
f
4)
))
(
(
x
f
g
)
.
Tenglikdan foydalanib,
f
(
x
)
ni toping (
Dostları ilə paylaş: